Question

A hydrocarbon `(A)` was found to have vapour density `36`. If forms only a single monochloro substitution product. Suggest `(A)`.

Answer 1

Vapour density `=36`, Molecular weight `= 2xx36= 72`
Examination of molecular weight suggests it to be an alkane, i.e., `C_(n)H_(2n+2)`.
Therefore, alkane `(A)` is `C_(5)H_(12)`
(A) gives only one product on substitution, thus `(A)` is:
`CH_(3)-underset(CH_(3))underset(|)overset(CH_(3))overset(|)(C )-CH_(3)overset(Cl_(2))rarrCH_(3)-underset(CH_(3))underset(|)overset(CH_(3))overset(|)(C )-CH_(2)Cl`
`(A)` is `(CH_(3))_(4)C` i.e., `2,2-` dimethyl propane (neopentane)